Question

Solve the following equation :
${\log }_{10}\left(3^x-2^{4-x}\right)=2+\frac{1}{4}{\log }_{10}16-\frac{x{\log }_{10}4}{2}$

Answer 1

${\log }_{10}(3^x-2^{4-x})=2+\frac{1}{4}{\log }_{10}16-\frac{x{\log }_{10}4}{2}$

$\Rightarrow {\log }_x(3^x-2^{4-x})=2+{\log }_{10}2-{\log }_{10}{2}^x$

$\Rightarrow {\log }_{10}(3^x-2^{4-x})+{\log }_{10}{2}^x=2+{\log }_{10}2$

$\Rightarrow {\log }_{10}(3^x-2^{4-x})\left(2^x\right)={\log }_{10}2=2$

$\Rightarrow {\log }_{10}\left(\frac{6^x-2^4}{2}\right)=100$

$\Rightarrow \frac{(6^x-2^4)}{2}=100$

$\Rightarrow 6^x-2^4=200$

$\Rightarrow 6^x=216$

$\Rightarrow x=3$